全A了，提供下项链O(N)解法，代码可以更精简点，过了就没管了

#include <iostream>
#include <string>
#include <deque>
using namespace std;
int cut(const string &neck)
{
	int i = 0;
	int j = 0;
	int mini = neck.length() / 2;
	int flag[5] = {0};
	int count_f = 5;
	deque<char> deq;
	deque<int> pos;
	while(i < neck.length() / 2 && j < neck.length())
	{
		if(neck[j] <= 'E')
		{
			if(deq.empty())
			{
				i = j;
				deq.push_back(neck[j]);
				pos.push_back(j);
				flag[neck[j] - 'A'] = 1;
				count_f--;
			}
			else if(deq.front() == neck[j])
			{
				deq.push_back(neck[j]);
				pos.push_back(j);
				flag[neck[j] - 'A']++;
				char head = deq.front();
				while(flag[head - 'A'] > 1)
				{
					deq.pop_front();
					pos.pop_front();
					flag[head - 'A']--;
					head = deq.front();
					i = pos.front();
				};
			}
			else
			{
				deq.push_back(neck[j]);
				pos.push_back(j);
				if(flag[neck[j] - 'A'] == 0)
				{
					count_f--;
				}
				flag[neck[j] - 'A']++;
			}
			if(count_f == 0)
			{
				if(mini > j - i + 1)
				{
					mini = j - i + 1;
				}
			}
		}
		j++;
	}
	return neck.length() / 2 - mini;
}
int main()
{
	string neck;
	while(cin>>neck)
	{
		cout<<cut(neck + neck)<<endl;
	}
	return 0;
}